Verified Solution • 2025–26 Session moderate

QJEEM19Mabdcf: JEE Main 2019 Mathematics - Class 12 JEEM19Mabdcf Solution: Let f(x) = 5 –

Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x ∈\in R. If f(x) attains maximum value at α\alpha and g(x) attains minimum value at β\beta , then lim⁡x→−αβ(x−1)(x2−5x+6)x2−6x+8\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}} is equal to :

Multiple Choice Options

  • A 12{1 \over 2}
  • B −12-{1 \over 2}
  • C 32{3 \over 2}
  • D −32-{3 \over 2}

Step-by-Step Verified Solution

From f(x) = 5 - | x - 2 |
maximum value of f(x) is at x = 2

From g(x) = | x + 1 |
minimum value of g(x) is at x = -1

∴\therefore αβ\alpha \beta = - 2

⇒\Rightarrow lim⁡x→2(x−1)(x−2)(x−3)(x−2)(x−4)\mathop {\lim }\limits_{x \to 2} {{(x - 1)(x - 2)(x - 3)} \over {(x - 2)(x - 4)}}

⇒\Rightarrow (2−1)(2−3)(2−4)=12{{(2 - 1)(2 - 3)} \over {(2 - 4)}} = {1 \over 2}

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